3-4-5 Triangle by a Kid

It is often said that the practice of hanging sangaku was so common that even women and children took part in that activity. While there is a reasonable doubt as to the popularity of sangaku, there is no question that some were hung by very young children. The one below was proposed by Sato Naosue, age thirteen, and hung in 1847 in the Akahagi Kannon temple in Ichinoseki city.

Two pink circles of radius r and two white circles of radius t are inscribed in a square, as shown. The square itself is inscribed in a large triangle and, as illustrated, two circles of radii r and R are inscribed in the small triangles outside the square. Show that R=2t.

Solution

References

1. H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008, pp. 104-105

Sangaku

1. Sangaku: Reflections on the Phenomenon
2. Critique of My View and a Response
3. 1 + 27 = 12 + 16 Sangaku
4. 3-4-5 Triangle by a Kid
5. 7 = 2 + 5 Sangaku
6. A 49^th Degree Challenge
7. A Geometric Mean Sangaku
8. A Hard but Important Sangaku
9. A Restored Sangaku Problem
   • A better solution to a difficult sangaku problem
   • A Trigonometric Solution to a Difficult Sangaku Problem
10. A Sangaku: Two Unrelated Circles
11. A Sangaku by a Teen
12. A Sangaku Follow-Up on an Archimedes' Lemma
13. A Sangaku with an Egyptian Attachment
14. A Sangaku with Many Circles and Some
15. A Sushi Morsel
16. An Old Japanese Theorem
17. Archimedes Twins in the Edo Period
18. Arithmetic Mean Sangaku
19. Bottema Shatters Japan's Seclusion
20. Chain of Circles on a Chord
21. Circles and Semicircles in Rectangle
22. Circles in a Circular Segment
23. Circles Lined on the Legs of a Right Triangle
24. Equal Incircles Theorem
    • Equal Incircle Theorem, Angela Drei's Proof
25. Equilateral Triangle, Straight Line and Tangent Circles
26. Equilateral Triangles and Incircles in a Square
27. Five Incircles in a Square
28. Four Hinged Squares
29. Four Incircles in Equilateral Triangle
    • Four Incircles in an Equilateral Triangle, a Sangaku
30. Gion Shrine Problem
31. Harmonic Mean Sangaku
32. Heron's Problem
33. In the Wasan Spirit
34. Incenters in Cyclic Quadrilateral
35. Japanese Art and Mathematics
36. Malfatti's Problem
37. Maximal Properties of the Pythagorean Relation
38. Neuberg Sangaku
39. Out of Pentagon Sangaku
40. Peacock Tail Sangaku
41. Pentagon Proportions Sangaku
42. Proportions in Square
43. Pythagoras and Vecten Break Japan's Isolation
44. Radius of a Circle by Paper Folding
45. Review of Sacred Mathematics
46. Sangaku à la V. Thebault
47. Sangaku and The Egyptian Triangle
48. Sangaku in a Square
49. Sangaku Iterations, Is it Wasan?
50. Sangaku with 8 Circles
51. Sangaku with Angle between a Tangent and a Chord
52. Sangaku with Quadratic Optimization
53. Sangaku with Three Mixtilinear Circles
54. Sangaku with Versines
55. Sangakus with a Mixtilinear Circle
56. Sequences of Touching Circles
57. Square and Circle in a Gothic Cupola
58. Steiner's Sangaku
59. Tangent Circles and an Isosceles Triangle
60. The Squinting Eyes Theorem
61. Three Incircles In a Right Triangle
62. Three Squares and Two Ellipses
63. Three Tangent Circles Sangaku
64. Triangles, Squares and Areas from Temple Geometry
65. Two Arbelos, Two Chains
66. Two Circles in an Angle
67. Two Sangaku with Equal Incircles

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Copyright ? 1996-2014 Alexander Bogomolny

Obviously, the side of the square is of length 4r. One application of the Pythagorean theorem (in the dashed triangle) shows that r=3t/2. Use the Pythagorean theorem again in the small upper triangle:

Let a be the small length of that triangle and c its hypotenuse. On the other hand, the area S of the small triangle can be computed in two ways:

2S=r(a+c+4r)=4r?a,

such that a+c+4r=4a; from which c=3a?4r. Substitute that into the Pythagorean identity - a2+(4r)2=c2 - to get the equation

a2+16r2=(3a?4r)2,

solving which gives a=3r.

It may come as a surprise that the triangle turns out to be the famous 3-4-5 or Egyptian triangle with the short side equal to 3r. From the similarity of the three triangles, all of them have the proportions 3-4-5 which leads to 4r=3R. And finally, R=4r/3=2t.

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Copyright ? 1996-2014 Alexander Bogomolny